The magnetic field of the Earth at a certain location is...

The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of $50.0 \mu \mathrm{T} .$ A proton is moving horizontally toward the west in this field with a speed of $6.20 \times 10^{6} \mathrm{m} / \mathrm{s} .$ (a) What are the direction and magnitude of the magnetic force the field exerts on this charge? (b) What is the radius of the circular arc followed by this proton?

Key Concepts

Lorentz Force

The Lorentz force describes the force exerted on a charged particle moving in a magnetic field. It is given by the equation F = q(v × B), where q is the charge, v is the velocity vector, and B is the magnetic field vector. This relationship explains how the direction and magnitude of the force are determined by both the speed and direction of the particle and the magnetic field.

Vector Cross Product and the Right-Hand Rule

The cross product in the Lorentz force equation determines the force's direction. The right-hand rule is a practical tool used to find the direction of the resulting vector when two vectors are crossed. It involves orienting the fingers in the direction of the first vector (velocity) and curling them toward the second vector (magnetic field) so that the thumb points in the direction of the magnetic force.

Circular Motion of Charged Particles

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force always acts perpendicular to its velocity, resulting in circular motion. This situation is governed by the balance between the magnetic force and the centripetal force needed to keep the particle in circular motion. The radius of this circular path is given by the formula r = mv/(qB), where m is the particle's mass, emphasizing the relationship between the physical properties of the particle and the magnetic environment.

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